The friction force between the box and the incline if the box does not slide down the incline will be 0.577
The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.
Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle
We have to find the friction force between the box and the incline if the box does not slide down the incline
Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:
F₁ = F₂
μmgcosΘ = mgsinΘ
μ = (mgsinΘ)/(mgcosΘ)
μ = tanΘ
μ = 0.577
Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577
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Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
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Answer:
0.44m/s
Explanation:
drift velocity=I/nAq
diameter 12 gauge
wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*
V = 0.44m/s
Answer:
The acceleration of the horse during this time interval is 24.286 m/min²
Explanation:
Given;
initial velocity of the horse during the gallop, u = 543 m /min
final velocity of the horse during the gallop, v = 628 m /min
time of motion, t = 3.5 minutes
The acceleration of the horse is given by the change in velocity per change in time;
Therefore, the acceleration of the horse during this time interval is 24.286 m/min²
